Asymptotic behavior of $\sum\limits_{k=1}^{n}k^{1/4}$

What is the asymptotic behavior of the sequence: $$s_n=\sum_{k=1}^{n}k^{1/4}$$ when $n\to \infty$?

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Are you familiar with Riemann sums? These yield immediately the equivalent $\frac45n^{5/4}$ (and even, much more precise inequalities). –  Did Jan 31 '12 at 11:41
Thank you. I've read the wikipedia link and now I think you mean something like this: looking at the interval $[0,n]$, subdivide it in $n$ subintervals. Then, $\sum^{n-1}k^{1/4}<\int_0^n x^{1/4} dx<\sum^n k^{1/4}$. The integral is just what you write. Is this correct? Can it be made more formal? And how can you get even more precise inequalities? Thanks again –  quark1245 Jan 31 '12 at 11:56
Pretty much. See answer. –  Did Jan 31 '12 at 12:40

Let $k\geqslant1$. Since the function $x\mapsto x^{1/4}$ is increasing, $(k-1)^{1/4}\leqslant x^{1/4}\leqslant k^{1/4}$ for every $k-1\leqslant x\leqslant k$. Integrating this double inequality yields $$(k-1)^{1/4}\leqslant\int_{k-1}^kx^{1/4}\mathrm dx\leqslant k^{1/4}.$$ Summing these from $k=1$ to $k=n$ yields $$s_{n-1}=\sum_{k=0}^{n-1}k^{1/4}\leqslant\int_{0}^nx^{1/4}\mathrm dx\leqslant \sum_{k=1}^{n}k^{1/4}=s_n.$$ Since the integral is $\tfrac45n^{5/4}$ and $s_{n-1}=s_n-n^{1/4}$, this yields, for every $n\geqslant1$, $$\tfrac45n^{5/4}\leqslant s_n\leqslant\tfrac45n^{5/4}+n^{1/4}.$$ Note finally that $n^{1/4}=o(n^{5/4})$ hence a (much weakened) version of this is $s_n\sim\tfrac45n^{5/4}$.

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Here is a diagram to accompany Did's fine answer:

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+1. Nice...  –  Did Jan 31 '12 at 17:45
How was the diagram made? –  lhf Feb 14 '12 at 16:11
@lhf Using JSXGraph. –  David Mitra Feb 14 '12 at 16:18
Nice, thanks a lot. –  lhf Feb 14 '12 at 16:19

Euler McLaurin Summation yields:

$$\sum_{k=0}^{n} k^{1/4} = \frac{4}{5} n^{5/4} + \frac{1}{2} n^{1/4} + C + \mathcal{O}(n^{-3/4})$$

It can be shown that $C = \zeta(-1/4)$ where $\zeta$ is the Riemann-Zeta function as defined on the whole plane.

Note that this is a much stronger statement than $\sum_{k=0}^{n}k^{1/4} \sim \frac{4}{5} n^{5/4}$, which (in the current context) means that

$$\lim_{n \to \infty} \frac{\sum_{k=0}^{n}k^{1/4}}{\frac{4}{5} n^{5/4}} = 1$$

What Euler Mclaurin gives us is the following:

$$\lim_{n \to \infty} \sum_{k=0}^{n}k^{1/4}- \frac{4}{5} n^{5/4} - \frac{1}{2} n^{1/4} - C = 0$$

Which actually implies that

$$\lim_{n \to \infty} \sum_{k=0}^{n}k^{1/4}- \frac{4}{5} n^{5/4} = \infty$$ There is also an elementary proof of this fact (but which does not determine the exact value of $C$) here: How closely can we estimate $\sum_{i=0}^n \sqrt{i}$

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